Chapter 8, Problem 8.10P

Interpretation Introduction

Interpretation:

The strain-hardening exponent for the metal needs to be determined. Whether the metal is FCC, BCC or HCP needs to be explained.

Concept introduction:

Strain hardening is called work hardening. Strengthening occurs due to plastic deformation. Strengthening occurs due to dislocation movement.

Answer to Problem 8.10P

Strain hardening exponent for the metal 'n' is 0.12. The value of strain hardening is in BCC metals.

Explanation of Solution

Given Information:

The measurements in the plastic regions are as follows:

Force (1b)	Change in gage length (in)   $(\Delta 1)$	Diameter (in)
27,500	0.2103	0.4800
27,000	0.4428	0.4566
25,700	0.6997	0.4343

Calculation:

Now, stress acting on the metal bar at 27500 lb

  ${\sigma }_1=\frac{f{d}_1}{A_f}$ --------------- (1)

But

  $A_f=\frac{\pi }{F}d{i}^2=\text{area}\text{of}\text{bar}$

Where,

  $\begin{array}{l}F{d}_1=\text{Force of metal bar}=27500lb\hfill \\ d_1=\text{Diameter of bar}=0.\text{48}\text{in}\text{.}\hfill \end{array}$

Equation (1) becomes,

  $\begin{array}{l}{\sigma }_1=\frac{27500}{\frac{\pi }{4}\times {0.48}^2}\\ {\sigma }_1=151970Psi\end{array}$

Now, stress acting at

  $\begin{array}{l}k{d}_2=27000lb,\\ d_2=0.4566in\\ {\sigma }_2=\frac{f{d}_2}{A_2}\\ {\sigma }_2=\frac{f{d}_2}{\frac{\pi }{4}{d}_2{}^2}\\ {\sigma }_2=\frac{27000}{\frac{\pi }{4}\times {0.4566}^2}\\ {\sigma }_2=164893Psi\end{array}$

At,

  $\begin{array}{l}k{d}_3=27000lb,\\ d_3=0.4343in\\ {\sigma }_3=\frac{257000}{\frac{\pi }{4}\times {0.4343}^2}\\ {\sigma }_3=173486Psi\end{array}$

Calculating 'n' for force 27500 lb act on metal by using formula.

  $\begin{array}{l}{\sigma }_1=k{\xi }_1{}^n\\ \ln {\sigma }_1\ln k+n\ln \xi \end{array}$ --------------- (2)

Where,

  $\begin{array}{l}\xi =\mathrm{true}strain=0.1\\ {\sigma }_1=27500lb\end{array}$

Equation (2) becomes,

  $\begin{array}{l}\ln (151970)=\ln k+n\ln (0.1)\\ 11.9314=\ln k+n(2.3026)\end{array}$ --------------------- (a)

Now,the force of 25700 lb act on metal, strain hardening is given by,

  ${\sigma }_3=k{\xi }_3{}^n$

Applying logarithm on both sides

  $\ln {\sigma }_3\ln k+n\ln \xi$

Here,

  $\begin{array}{l}\xi =\text{True strain}=0.3\hfill \\ {\sigma }_3=\text{Stress acting}=25700lb \hfill \end{array}$

Equation (3) becomes,

  $\begin{array}{l}\ln (173486)=\ln k+n\ln (0.3)\\ 12.0636=\ln k+n(1.2040)\end{array}$ ---------------- (b)

From equation (a) and (b),

  $n=0.12$

Conclusion

Thus, strain hardening exponent for metal 0.12 which is in BCC range.